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Step-by-Step Solution
Step 1: Rewrite the given circle in standard form
The given circle is
$x^2 + y^2 - 4x + 6y - 12 = 0.$
Complete the square for $x$ and $y$:
$x^2 - 4x = (x - 2)^2 - 4,$
$y^2 + 6y = (y + 3)^2 - 9.$
Substituting these into the equation:
$(x - 2)^2 - 4 + (y + 3)^2 - 9 - 12 = 0,$
$(x - 2)^2 + (y + 3)^2 - 25 = 0,$
or
$(x - 2)^2 + (y + 3)^2 = 25.$
Hence, the center of this circle is $C_1 = (2,\,-3)$ and its radius is $r_1 = 5.$
Step 2: Identify the center of the second circle
The problem states that the second circle $S$ has its center at
$C_2 = (-3,\,2).$ We need to find the radius $r_2$ of this circle.
Step 3: Understand the diameter-chord condition
A diameter of the first circle (center $C_1$, radius $5$) is a chord of the second circle (center $C_2$, unknown radius $r_2$).
If $A$ and $B$ are the endpoints of that diameter on the first circle:
$A$ and $B$ lie on the circle $(x - 2)^2 + (y + 3)^2 = 25.$
$C_1$ is the midpoint of $AB.$
The chord $AB$ must lie on the second circle with center $C_2$ and radius $r_2.$
Because $AB$ is a diameter of the first circle, $|AB| = 2 \times 5 = 10.$
Step 4: Parametrize the endpoints of the diameter
Let $A$ and $B$ be such that $C_1$ is their midpoint. If we set
$A = (2 + h,\,-3 + k)$ and $B = (2 - h,\,-3 - k),$
then $C_1 = \big(\tfrac{(2+h) + (2-h)}{2},\, \tfrac{(-3+k) + (-3-k)}{2}\big) = (2,\,-3).$
Since $A$ and $B$ are endpoints of a diameter of circle 1, the distance $|A - B| = 10.$
Observe
\[
|A - B| = 2\sqrt{h^2 + k^2} \quad \Longrightarrow \quad 2\sqrt{h^2 + k^2} = 10 \quad \Longrightarrow \quad h^2 + k^2 = 25.
\]
Step 5: Impose that $A$ and $B$ also lie on circle $S$
For $A$ to lie on circle $S$ (center $C_2=(-3,\,2)$, radius $r_2$), we need
\[
|A - C_2|^2 = r_2^2.
\]
Compute
\[
A - C_2 = ( (2 + h) - (-3),\, (-3 + k) - 2 ) = (5 + h,\, k - 5),
\]
\[
|A - C_2|^2 = (5 + h)^2 + (k - 5)^2.
\]
Since $h^2 + k^2 = 25,$ expand and simplify:
\[
(5 + h)^2 + (k - 5)^2 = (25 + 10h + h^2) + (k^2 - 10k + 25)
= (h^2 + k^2) + 10h - 10k + 50
= 25 + 10(h - k) + 50
= 75 + 10(h - k).
\]
Similarly, for $B = (2 - h,\,-3 - k),$
\[
B - C_2 = ( (2 - h) - (-3),\, (-3 - k) - 2 ) = (5 - h,\, -5 - k),
\]
\[
|B - C_2|^2 = (5 - h)^2 + (-5 - k)^2
= (25 - 10h + h^2) + (25 + 10k + k^2)
= (h^2 + k^2) - 10h + 10k + 50
= 25 - 10(h - k) + 50
= 75 - 10(h - k).
\]
Both $A$ and $B$ must be on the same circle $S$, so
\[
|A - C_2|^2 = |B - C_2|^2 = r_2^2.
\]
Thus,
\[
75 + 10(h - k) = 75 - 10(h - k).
\]
This forces
\[
10(h - k) = -10(h - k) \quad \Longrightarrow \quad h - k = 0 \quad \Longrightarrow \quad h = k.
\]
Step 6: Solve for $r_2$
Since $h^2 + k^2 = 25$ and $h = k,$ we get
\[
2h^2 = 25 \quad \Longrightarrow \quad h^2 = \frac{25}{2} \quad \Longrightarrow \quad h = \pm \frac{5}{\sqrt{2}}, \; k = \pm \frac{5}{\sqrt{2}}.
\]
Substituting $h = k$ into
$|A - C_2|^2 = 75 + 10(h - k)$
shows
\[
h - k = 0 \quad\Rightarrow\quad 75 + 10(h - k) = 75.
\]
Therefore,
\[
r_2^2 = 75 \quad \Longrightarrow \quad r_2 = \sqrt{75} = 5\sqrt{3}.
\]
Final Answer
The radius of the circle $S$ is $5\sqrt{3}$.
